155. Min Stack

Medium (Trung bình) Python 🔗 Xem trên LeetCode

📋 Đề Bài

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

Implement the MinStack class:

  • MinStack() initializes the stack object.
  • void push(int val) pushes the element val onto the stack.
  • void pop() removes the element on the top of the stack.
  • int top() gets the top element of the stack.
  • int getMin() retrieves the minimum element in the stack.

You must implement a solution with O(1) time complexity for each function.

 

Example 1:

Input
["MinStack","push","push","push","getMin","pop","top","getMin"]
[[],[-2],[0],[-3],[],[],[],[]]

Output
[null,null,null,null,-3,null,0,-2]

Explanation
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); // return -3
minStack.pop();
minStack.top();    // return 0
minStack.getMin(); // return -2

 

Constraints:

  • -231 <= val <= 231 - 1
  • Methods pop, top and getMin operations will always be called on non-empty stacks.
  • At most 3 * 104 calls will be made to push, pop, top, and getMin.

🧠 Thuật Toán & Kỹ Thuật

Stack (Ngăn xếp)
⏱️ Thời gian O(n)
💾 Không gian O(1)

💻 Lời Giải

Python 0155-min-stack.py 
class MinStack:

    def __init__(self):
        self.stackNormal = []
        self.stackMinVal = []

    def push(self, val: int) -> None:
        self.stackNormal.append(val)
        if len(self.stackMinVal) == 0 or val <= self.stackMinVal[-1]:
            self.stackMinVal.append(val)

    def pop(self) -> None:
        if self.stackMinVal[-1] == self.stackNormal[-1]:
            self.stackMinVal.pop()
        self.stackNormal.pop()

    def top(self) -> int:
        return self.stackNormal[-1]

    def getMin(self) -> int:
        return self.stackMinVal[-1]


# Your MinStack object will be instantiated and called as such:
# obj = MinStack()
# obj.push(val)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.getMin()